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In quantum mechanics, the definition of symmetry breaking is nontrivial. See What is spontaneous symmetry breaking in QUANTUM systems?

Let me briefly summarize that question:

  1. In spin-$1/2$ quantum ferromagnetic Heisenberg model: $$H_1=-\sum_{i}\hat{\mathbf{S}}_i\cdot \hat{\mathbf{S}}_{i+1}$$ the exact ground state of finite size or infinite size is all spin point out in the same direction, e.g. $|\uparrow \cdots\rangle$,$|\downarrow \cdots\rangle$. But in QM, the superposition of this two state is still the ground state.

  2. In some model like transverse Ising model(for $|h|<1$) $$H_2=\sum_i(-\sigma^z_i \sigma^z_{i+1}+h \sigma^x_i)$$ For finite system, the ground state is unique and doesn't break the $\mathbb{Z}_2$ symmetry. For infinite system, the ground state is double degenerate.

So the usual way to define spontaneous symmetry breaking (SSB),e.g. ground state has lower symmetry than the system, seems to be ill-defined. For quantum ferromagnetic Heisenberg model, there exist symmetric ground state without breaking any symmetry no matter in finite or inifinite size of system. In transverse Ising model, for any finite size of system, the ground state is even unique and does not break the $\mathbb{Z}_2$ symmtry. Even in infinite size, there still exist the symmetric ground state.

Prof. Wen gave an unambiguous definition of spontaneous symmetry breaking in quantum system.

Definition(Wen): A model is called spontaneous symmetry breaking(SSB) if there exist a symmetric ground state which is GHZ state.

No matter in a system with or without SSB, there always exist symmetric ground state as we see from above example. But symmetric ground stete is unstable(GHZ type) in SSB system.

My questions

  1. How to rigorously argue that the superposition state is unstable in spontaneously spontaneous symmetry breaking case.(I think I should have relation with decoherence.)

I heard following different kinds of explainations which I hardly understand:

The first saying is that SSB can only occur in infinite large system because the tunneling between different degenerate vacuum is exponentially damping as the size of system.

  1. In ferromagnetic Heisenberg model, $\langle\downarrow \cdots |H_1 |\uparrow \cdots\rangle$ is always $0$ no matter the system is finite or infinite. But we know in finite size ferromagnetic Heisenberg model can have superposition state. It seems that "tunneling amplitude is zero" has no relationship with "stability of symmetric state".

  2. Furthermore, how can degenerate ground states have tunneling amplitude? Because if there is tunneling between different degenerate ground states, there exist off-diagonal term, then they are not the ground state.

For example, $$H =\begin{bmatrix}1 & 0\\0&1 \end{bmatrix}$$ If $(1,0)$ and $(0,1)$ have tunneling amplitude, it means Hamitonian is $$H' =\begin{bmatrix}1 & \epsilon \\\epsilon &1 \end{bmatrix}$$ Then $(1,0)$ and $(0,1)$ are not ground state.

The second saying is that in SSB model, under the perturbation breaking the symmetry, the symmetric state is exponentially unstable as the size go to infinite.

  1. What's the meaning of this sentence? Because whether the model is spontaneous symmetry breaking or not, if you add a symmetry breaking term to original Hamitonian, the ground state always breaks the symmetry.

For example, transverse Ising model with $h>1$, the ground state don't break the symmetry. If you add a perturbation term $\sum_i t \sigma^z_i$ to this Hamitonian, the ground state always break the $\mathbb{Z}_2$ symmetry.

$$H_3=\sum_i(-\sigma^z_i \sigma^z_{i+1}+2 \sigma^x_i +t \sigma^z_i)$$ No matter how small of $t>0$, the ground state of $H_3$ always breaks the the $\mathbb{Z}_2$ symmetry.

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    $\begingroup$ Could it not be as simple as the fact that a measurement of any single spin will collapse the superposition? That fact would not be true for the pure all up and all down spin states, which agrees with the notion that such states are stable $\endgroup$ – KF Gauss Dec 12 '17 at 5:03
  • $\begingroup$ @user157879 But if you're true, the superposition state can't exist for even two spins. $\endgroup$ – maplemaple Dec 12 '17 at 7:13
  • $\begingroup$ well if you have a superposition of two spins and one of them (50% of the total number of spins!) is interacting with outside states, then yes the superposition will break. Usually one would assume that states will only interact with some small probability, but with a macroscopic number of states, the probability of a single one interacting will then be basically 1 $\endgroup$ – KF Gauss Dec 12 '17 at 7:31
  • $\begingroup$ To the finite ferromagnetic Heisenberg model : The ground states are the states of maximal total Spin. Therefore each ground state breaks rotational symmetry (otherwise its total spin would be zero) . $\endgroup$ – jjcale Dec 12 '17 at 17:27
  • $\begingroup$ @jjcale Yes, they are one basis for the space of degenerate ground state. But you can always add them together to form a symmetric ground state which do not break the symmetry. $\endgroup$ – maplemaple Dec 13 '17 at 23:03
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1.

The key point that you're missing is that spontaneous symmetry breaking, or indeed the notion of phase transitions in general, only works for systems with local interactions. A phase transition is defined to be a point in Hamiltonian parameter space at which the free-energy density becomes non-analytic in the infinite-size limit. This definition obviously presupposes the existence of a well-defined infinite-system limit of the free energy density. But for translationally invariant lattice systems like the Ising model, the free energy density only approaches a constant value as $N \to \infty$ if $\sum_j J_{ij}$ converges absolutely, which roughly means that $|J_{ij}|$ has to fall off faster than $1/r^d$, where $d$ is the number of dimensions. In other words, the couplings must be reasonably local.

(Experts might object that disordered systems with nonlocal all-to-all couplings, like the Sherrington-Kirkpatrick or SYK models, still have replica-symmetry-breaking phase transitions. But that's actually only true if you rescale the coupling constants as a power of the total system size, which is not a very physical thing to do. If you don't do this, then the phase transition goes away, and indeed the $N \to \infty$ limit becomes ill-defined. Real systems are never truly all-to-all coupled - in practice there's some maximum distance at which the couplings go away, and all-to-all-coupled models are just a convenient approximation.)

Any putative explanation of spontaneous symmetry breaking that doesn't explicitly use locality is at best seriously incomplete. Decoherence is too complicated to explain here, but a key assumption is that the interactions are local in space, which picks out the position basis as a naturally favored pointer basis, so that position near-eigenstates are more natural than, say, momentum near-eigenstates.

  1. and 3.

The locality of the system, and specifically the assumption that perturbations are all local, gives us a notion of the "distance" between two states which is more useful than mere othogonality. As you point out, orthogonality/inner products alone can't distingush between two states that only differ by a single spin, and two states that differ by all their spins, even though the latter pair is clearly in some sense "more different" than the former.

You are of course correct that $\langle i | A | j \rangle = 0$ for any two distinct eigenstates of any Hermitian operator, not just the Hamiltonian. But that simple matrix element isn't the right definition of "the tunneling amplitude". As far as I know, the actual definition is a little fuzzy and the concept is more of an art than a science, but here are two possible conceptualizaions:

a) You can think of the symmetry-breaking term as a perturbation and decompose the Hamiltonian as $H = H_0 + \Delta H$, where $H_0$ respects the symmetry and $\Delta H$ breaks it. Then perturbation theory tells us that all the perturbative corrections can be expressed in terms of the matrix elements $\langle i_0 | \Delta H | j_0 \rangle$ where $\langle i_0|$ and $|j_0\rangle$ are the eigenstates of the unperturbed Hamiltonian $H_0$, not the exact Hamiltonian. These matrix elements are generically nonzero.

b) I don't like perturbation theory, so I prefer to think of it by analogy with Monte Carlo. The environment is constantly try to act on the system with random little local symmetry-breaking perturbations. You can think of it as if $h = 0$ in the full Hamiltonian, but $h_i \sigma_i^x$ terms randomly appear momentarily at individual sites $i$ (or similar terms on small local clusters of sites). These are like Monte Carlo candidate spin flips, and at low temperature they usually only get accepted if they lower the system's total energy. For a small system that starts in the all-$\uparrow$ state, you might get lucky and accept enough flips to eventually take you into a majority-$\downarrow$ state, at which point you'll then probably proceed to all-$\downarrow$ - even though each of those first few individual flips were unlikely. But in order to flip more than half the system, you initially need to get lucky many (independent) times in a row, and the odds of that happening decrease exponentially with system size. The "tunneling amplitude" is basically the probability of this happening after many Monte Carlo sweeps, and it indeed decreases exponentially with system size. For a small system, you'll eventually flip over to the other ground state, although it'll take a really long time. For a large system, it'll take a really long time, and for an infinite system it'll never fully get there.

If that analogy's too classical for your taste, you can instead think of the space of random quantum circuits, with circuits weighted according to a cost function that depends on the Hamiltonian matrix elements, and the "tunneling amplitude" between two quantum states is like the total weight of all the random circuits that take one state to the other.

4.

You're right that any finite value of $h$ breaks the symmetry. For any system, even an infinite one, you have that $m(h) \neq 0$ if $h > 0$. But what about the limit $h \to 0^+$? One definition of SSB is the failure of the limits $h \to 0^+$ and $N \to \infty$ to commute. In the SSB phase, after you take $N \to \infty$, you have that $m(h)$ has a jump discontinuity at $h = 0$, so that $m(0) = 0$ but $\lim \limits_{h \to 0^+} m(h) > 0$. That what we mean when we say that an infinitesimal perturbation $h$ breaks the symmetry.

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There are already some nice answers, but I'll give a physical, to-the-point one for those in a hurry.

The question I will address is

Why are cat states (such as GHZ states) physically unstable?

The answer being

Because a generic interaction with the environment---no matter how small---will collapse/decohere such states.

For example, consider the usual Ising chain where a single spin happens to interact with a single spin in the environment: $$ H = - \left( \sum_{n=1}^{N-1} \sigma^z_n \sigma^z_{n+1} \right) - \varepsilon \; \sigma^z_1 \sigma^z_\textrm{env} \qquad (\textrm{with }\varepsilon >0).$$

The cat state ground states will be $$ |\Psi_\pm \rangle = \frac{1}{\sqrt{2}}\left(|\uparrow_1 \uparrow_2 \cdots \uparrow_N \uparrow_\textrm{env} \rangle \pm |\downarrow_1 \downarrow_2 \cdots \downarrow_N \downarrow_\textrm{env} \rangle \right). $$ Supposing we don't have (coherent) control/knowledge about the environment, our effective description is $$ \rho_\textrm{system} = \textrm{Tr}_{\textrm{env}} \left( |\Psi_\pm\rangle \langle \Psi_\pm | \right) = \frac{1}{2} \left( \rho_+ + \rho_- \right),$$ where $\rho_\pm$ are the symmetry-broken states of the system (corresponding to $|\uparrow_1 \cdots \uparrow_N\rangle$ and $|\downarrow_1\cdots \downarrow_N\rangle$).


NB: The above also explains why the GHZ states appearing as the ground states of topological fermionic chains (e.g. the Kitaev chain) are stable: there is no local operator that splits up the two-dimensional Hilbert space, whereas the above worked cause there was such a local operator (i.e. $\sigma^z_1$) to which the environment could couple (the latter functioning as a measurement apparatus).

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    $\begingroup$ Are you saying that the ground state of the Kitaev chain is a coherent superposition of states of different fermion parity? $\endgroup$ – Lorenz Mayer Nov 13 '18 at 9:51
  • $\begingroup$ @LorenzMayer I said that there was no local operator that could couple to fermionic parity: that is true, but it indeed gave the wrong implication (i.e. the one you said), my bad. Fixed it :) Thanks $\endgroup$ – Ruben Verresen Nov 14 '18 at 1:22
  • $\begingroup$ How are you defining GHZ states? I would define them as states that violate the cluster decomposition property, but the ground states of the Kitaev chain respect that property so I wouldn't call them GHZ states. $\endgroup$ – tparker Nov 18 '18 at 0:38
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The problem is that, talking for example about the transverse field Ising model, the states $\Omega_\pm$ with all spins having spin-$z$-eigenvalue $\pm 1$ do not exist in the same Hilbert space in the thermodynamic limit.

This seems like a weird statement, so i will try to explain it further. Working directly in the thermodynamic limit, we should be careful with what operators we allow. Considering that measuring devices have only finite size, we should allow only for local operators, that is those operators which in some way decay at infinity (for example, exponentially, or with finite support).

Then for all such operators $O$, we have that

$$\langle \Omega_-, O \Omega_+ \rangle = 0.$$

That is, there is no operator taking you from $\Omega_-$ to $\Omega_+$. (This is possibly related to this "tunneling probability" business). This is however exactly the definition of a superselection sector! Hence what you thought is the unique symmetric ground state

$$\frac{\Omega_+ + \Omega_-}{\sqrt{2}}$$

is not a superposition, but actually a statistical mixture of two symmetry broken states.

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    $\begingroup$ I get what you're saying and it's a good point, but I think more is understood in problems like this one by identifying a proper metric of stability in the finite $N$ case and then taking the limit $N \to \infty$ at the end. $\endgroup$ – DanielSank Oct 8 '18 at 21:12
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    $\begingroup$ What happens to this explanation for the case of the Kitaev chain, where the cat state is the physical state? $\endgroup$ – Ruben Verresen Nov 13 '18 at 0:08

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